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what is the value of integration limit n-> infinity [n!/n to the power n]to the power 1/n
Solution) limit n-->inf. [1 + (n!-n^n)/n^n]^1/n
= e^ limit n-->inf. {(n!-n^n)/n^n}.1/n
(applying formula lim x-->a [1 + f(x)]^g(x) = e ^ {lim x-->a f(x).g(x)} )
=e ^ lim n-->inf. (n-1)!/(n)^n-1 - 1/n
=e ^ lim n---> inf. (1-1/n)(1-2/n)(1-3/n)......1/n - 1/n
=e ^ (1-o)(1-0).....(1-0).0 - 0
=e ^ 0-0
=1
hence, integation 1= x + C
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1+1
g/6-2+(9/9)
1+2i=a+ib so find a and b
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